Mathematics

# Find $\int \dfrac { d x } { x ^ { 2 } - a ^ { 2 } }$  and hence evaluate $\int \dfrac { d x } { x ^ { 2 } - 25 }$.

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{{x}^{2}-{a}^{2}}}$

$=\displaystyle\int{\dfrac{dx}{\left(x-a\right)\left(x+a\right)}}$

$=\dfrac{1}{2a}\displaystyle\int{\dfrac{\left(\left(x+a\right)-\left(x-a\right)\right)dx}{\left(x-a\right)\left(x+a\right)}}$

$=\dfrac{1}{2a}\displaystyle\int{\dfrac{\left(x+a\right)dx}{\left(x-a\right)\left(x+a\right)}}-\displaystyle\int{\dfrac{\left(x-a\right)dx}{\left(x-a\right)\left(x+a\right)}}$

$=\dfrac{1}{2a}\displaystyle\int{\dfrac{dx}{\left(x-a\right)}}-\displaystyle\int{\dfrac{dx}{\left(x+a\right)}}$

$=\dfrac{1}{2a}\left[\log{\left|x-a\right|}-\log{\left|x+a\right|}\right]+c$

$=\dfrac{1}{2a}\log{\left|\dfrac{x-a}{x+a}\right|}+c$

Now $\displaystyle\int{\dfrac{dx}{{x}^{2}-25}}$

$=\displaystyle\int{\dfrac{dx}{{x}^{2}-{5}^{2}}}$

$=\dfrac{1}{2\times 5}\log{\left|\dfrac{x-5}{x+5}\right|}+c$

$=\dfrac{1}{10}\log{\left|\dfrac{x-5}{x+5}\right|}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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